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If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?

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If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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New post 21 Aug 2018, 06:05
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Question Stats:

92% (01:20) correct 8% (02:53) wrong based on 48 sessions

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Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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New post 21 Aug 2018, 06:10
Taking LCM of X gives X = 7/9
Then (1-X)^2 = (1-7/9)^2 = 4/81
B is the answer

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If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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New post 21 Aug 2018, 11:32
Bunuel wrote:
If \(x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}\) then \((1−x)^2\) ?

A. 1/ 9
B. 4/81
C. 25/144
D. 9/16
E. 25/36


\(x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}\)

Or, \(x = \frac{28 - 30 + 30}{36}\)

Or, \(x = \frac{7}{9}\)

So, \((1−x)^2\)

= \((1 - \frac{7}{9})^2\)

= \((\frac{2}{9})^2\)

= \(\frac{4}{81}\), Answer must be (B)
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Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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New post 22 Aug 2018, 04:45
1
Bunuel wrote:
If \(x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}\) then \((1−x)^2\) ?

A. 1/ 9
B. 4/81
C. 25/144
D. 9/16
E. 25/36


Do not rush into calculations

\(x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}\)

\(\frac{15}{18}\)=\(\frac{5}{6}\)

\(\frac{10}{12}\)=\(\frac{5}{6}\)

So, \(x = \frac{7}{9} − \frac{5}{6} + \frac{5}{6}\) = \(\frac{7}{9}\)

\((1−x)^2\) = \((1−\frac{7}{9})^2\) = \(\frac{4}{81}\)

Answer: B
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Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?   [#permalink] 22 Aug 2018, 04:45
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